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Question Number 212844 by efronzo1 last updated on 25/Oct/24

$⇃ \underset{―}{}$
Commented by AlagaIbile last updated on 29/Oct/24

$d e t (S^{2}) = {d e t}^{2} (S) = 36$ $d e t (S) = \pm 6$ $B y C - H T h e o r e m :$ $S^{2} - t r (S) S + (\pm 6) = 0$ $I n t r o d u c e t r a c e t o d e q n$ $t r (S^{2}) - {t r}^{2} (S) + 2 (\pm 6) = 0$ $13 - {t r}^{2} (S) \pm 12 = 0$ $t r (S) = \pm 5, \pm 1$ $F o r t r a c e (S) = \pm 5$ $∴ \pm 5 S = S^{2} + 6 I_{2}$ $S = \pm [\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}]$ $F o r t r a c e (S) = \pm 1$ $\pm S = S^{2} - 6 I_{2}$ $S = \pm [\begin{matrix} 8 & - 5 \\ 10 & - 7 \end{matrix}]$
	Answered by Rasheed.Sindhi last updated on 30/Oct/24
	$let S=± ((a,b),(c,d) ) ; assuming a,b,c,d∈Z S^2 =(± ((a,b),(c,d) ))^2 = (((a^2 +bc),(ab+bd)),((ca+cd),(bc+d^2 )) )= (((14 −5)),((10 −1)) ) a^2 +bc=14...i , b(a+d)=−5...ii c(a+d)=10...iii , bc+d^2 =−1...iv i − iv:a^2 −d^2 =15 { (((a−d)(a+d)=3×5⇒ { ((a−d=3)),((a+d=5)) :}⇒a=4,d=1)),(((a−d)(a+d)=5×3⇒ { ((a−d=5)),((a+d=3)) :}⇒a=4,d=−1)),(((a−d)(a+d)=1×15⇒ { ((a−d=1)),((a+d=15)) :}⇒a=8,d=7)),(((a−d)(a+d)=15×1⇒ { ((a−d=15)),((a+d=1)) :}⇒a=8,d=−7)) :} ii⇒ b=((−5)/(a+d)) iii⇒c=((10)/(a+d)) a=4,d=1⇒b=−1,c=2 a=4,d=−1⇒b=((−5)/3)∉Z , c=((10)/3)∉Z Rejected a=8,d=7⇒b=−(1/3)∉Z,c=((10)/(15))=(2/3)∉Z Rejected a=8,d=−7⇒b=−5,c=10 S=± ((4,(−1)),(2,( 1)) ) ,± ((8,(−5)),((10),(−7)) )$
	$l e t S = \pm (\begin{matrix} a & b \\ c & d \end{matrix}); a s s u m i n g a, b, c, d \in Z$ $S^{2} = {(\pm (\begin{matrix} a & b \\ c & d \end{matrix}))}^{2} = (\begin{matrix} a^{2} + b c & a b + b d \\ c a + c d & b c + d^{2} \end{matrix}) = (\begin{matrix} 14 - 5 \\ 10 - 1 \end{matrix})$ $a^{2} + b c = 14 . . . i, b (a + d) = - 5 . . . i i$ $c (a + d) = 10 . . . i i i, b c + d^{2} = - 1 . . . i v$ $i - i v : a^{2} - d^{2} = 15$ ${\begin{cases} (a - d) (a + d) = 3 \times 5 \Rightarrow {\begin{cases} a - d = 3 \\ a + d = 5 \end{cases} \Rightarrow a = 4, d = 1 \\ (a - d) (a + d) = 5 \times 3 \Rightarrow {\begin{cases} a - d = 5 \\ a + d = 3 \end{cases} \Rightarrow a = 4, d = - 1 \\ (a - d) (a + d) = 1 \times 15 \Rightarrow {\begin{cases} a - d = 1 \\ a + d = 15 \end{cases} \Rightarrow a = 8, d = 7 \\ (a - d) (a + d) = 15 \times 1 \Rightarrow {\begin{cases} a - d = 15 \\ a + d = 1 \end{cases} \Rightarrow a = 8, d = - 7 \end{cases}$ $i i \Rightarrow b = \frac{- 5}{a + d} i i i \Rightarrow c = \frac{10}{a + d}$ $a = 4, d = 1 \Rightarrow b = - 1, c = 2$ $a = 4, d = - 1 \Rightarrow b = \frac{- 5}{3} \notin Z, c = \frac{10}{3} \notin Z R e j e c t e d$ $a = 8, d = 7 \Rightarrow b = - \frac{1}{3} \notin Z, c = \frac{10}{15} = \frac{2}{3} \notin Z R e j e c t e d$ $a = 8, d = - 7 \Rightarrow b = - 5, c = 10$ $S = \pm (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}), \pm (\begin{matrix} 8 & - 5 \\ 10 & - 7 \end{matrix})$
	Commented by Ghisom last updated on 30/Oct/24
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	Answered by Rasheed.Sindhi last updated on 26/Oct/24

	$l e t S = (\begin{matrix} a & b \\ c & d \end{matrix})$ $S^{2} = {(\begin{matrix} a & b \\ c & d \end{matrix})}^{2} = (\begin{matrix} a^{2} + b c & a b + b d \\ c a + c d & b c + d^{2} \end{matrix}) = (\begin{matrix} 14 - 5 \\ 10 - 1 \end{matrix})$ $a^{2} + b c = 14, b (a + d) = - 5$ $c (a + d) = 10, b c + d^{2} = - 1$ $a + d = - \frac{5}{b} = \frac{10}{c} \Rightarrow c = - 2 b$ $b c = 14 - a^{2} = - 1 - d^{2} \Rightarrow a^{2} - d^{2} = 15$ $\Rightarrow (a - d) (a + d) = 3 \times 5$ $l e t a - d = 3$ $a + d = 5 \Rightarrow a = 5 - d$ $S = (\begin{matrix} 5 - d & b \\ - 2 b & d \end{matrix})$ $b = - 1, d = 1$ $O n e o f a n s w e r s :$ $S = (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix})$ $V e r i f i c a t i o n :$ $S^{2} = (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}) (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}) = (\begin{matrix} 16 - 2 & - 4 - 1 \\ 8 + 2 & - 2 + 1 \end{matrix}) = (\begin{matrix} 14 & - 5 \\ 10 & - 1 \end{matrix})$
	Commented by Rasheed.Sindhi last updated on 26/Oct/24

	$T h a n X s i r!$ $P l s e e m y n e x t a n s w e r, w h e r e$ $I c o v e r e d a l l i n t e g r a l s o l u t i o n s .$
	Commented by Ghisom last updated on 26/Oct/24

	$there are 4 solutions$ $\pm (\begin{matrix} 4 & - 1 \\ 2 & 1 \end{matrix}) and \pm (\begin{matrix} 8 & - 5 \\ 10 & - 7 \end{matrix})$
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